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anonymous
 one year ago
A car of mass 1.5 x 103 kg is hoisted on the large cylinder of a hydraulic press. The area of the large piston is 0.20 m2, and the area of the small piston is 0.015 m2.
a.Calculate the magnitude of the force of the small piston needed to raise the car with slow speed on the large piston.
b.Calculate the pressure, in pascals and kilopascals, in this hydraulic press.
anonymous
 one year ago
A car of mass 1.5 x 103 kg is hoisted on the large cylinder of a hydraulic press. The area of the large piston is 0.20 m2, and the area of the small piston is 0.015 m2. a.Calculate the magnitude of the force of the small piston needed to raise the car with slow speed on the large piston. b.Calculate the pressure, in pascals and kilopascals, in this hydraulic press.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hey there how are you?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I hope all is well with you;)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! I'm fine! thanks! :)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1here you have to keep in mind the principle of Pascal, namely the pressure of the fluid is the same on both pistons

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Where water is used as multiplier of force in call cases.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Since for question a I know the area of small piston

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I guess all I need is divide the mass to be lifted by the area of the piston?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1more precisely are used special oils, instead of water. the pressure exerted by the fluid on larger piston is: \[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}} = ...Pascals\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right... I needed to calculate the normal force acting on the object divided by the area of the piston.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0According to my calculator the pascals required on the piston is 73500 pascals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So that's considering if we only use the large piston right?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I got p = 73.6 kPa approximately

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On the other hand if I use the smaller piston the answer would be slightly greater I think

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got p=981000 on my side for the smaller piston.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, the pressure is the same through all the fluid. The force which is acting on the smaller piston, is perpendicular with respect to that piston, and its magnitude is: \[\Large F = p \times s = 73575 \cdot 0.015 = ...Newtons\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh ok.. but why are we only using the pressure applied to the large piston? Is it because of the structure of hydraulic press?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, we are using the pression of the fluid which is transmitted by the fluid itself, namely your hydraulic press can be represented as below: dw:1439828282269:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1what changes are the magnitude of the forces applied to both pistons

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So basically a force is applied to one end and that results in other end

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! we apply a little force and the hydraulic press multiply this little force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.073575 Pa on the large piston.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand until that point

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But then, isn't it like by 0.2/0.015 that the pressure is multiplied?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1that pressure is acting inside all the fluid, so the pressure of 73575 is acting on both pistons

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the pressure is the same!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you saying that fluid supplied from the inside the hydraulic press?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1in your exercise, we apply a force whose magnitude is 1,104 Newtons, and we can rise a car whose weight is: 14,715 Newtons

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! the fluid is working inside the hydraulic press

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The force the small piston needs to lift the car right? so I need to apply force on the small piston to lift the car to the larger piston and force multiplies

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1correct! we can see a hydraulic press as a multiplier of force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So it's correct to say 73575 Pa* 0.015= pressure require to lift the car? The only thing I don't understand is why we multiply by 0.015 for a pressure calculated using 0.2. Isn't it like we have to multiply by 0.015/0.2 ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1pressure is 73575 Pa, and 73575*0.015 is a force

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0likewise 73575 Pa*0.2 is the force I need to apply to larger piston to do the same trick?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1better is: 73575*0.2 is the force which is acting on the larger piston

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait a moment it's taking long so

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's my thanks for being patient with me XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am trying my best to understand

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You previously calculated the pressure applied to the large piston using the formula Mg/s=pascals. What if you used 0.015 instead of the 0.2? The overall pressure in the hydraulic press would then increase no?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0According to your reasoning that pressure is the same across all points in a hydraulic press, then the pressure calculated using 0.015 instead of 0.2 would be greater...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I have used 0.2 since 0.2 is the area of the piston on which the weight of the car is acting on

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I can not use 0.015, since it will be an error

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1dw:1439829785001:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1what I can know is the pressure transmitted by the fluid through all the hydraulic press, and that pressure, by definition, is: \[\Large p = \frac{{Mg}}{S} = \frac{{1500 \cdot 9.81}}{{0.2}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see then in terms of the larger piston the pressure acting on it is calculated by Mg/0.2. Right the pressure acting on the larger piston is accounted for by that. On the other hand, when I need to apply the pressure exerted by the car on larger piston, multiplying the previous pascals with 0.15 will be in accord with the principles of multiplication achieved by water.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So multiplying that number with 0.015 will give the force required to lift the car although at a slower rate.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Am I on the right line of reasoning?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! if I multiply the pressure by the area of the smaller piston, I get the force (magnitude) which I have to apply to that smaller piston, in order to raise the car

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1the concept is that pressure is the same in all points of the hydraulic press Whereas what is different are the magnitude of the forces which are acting on both pistons.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right.. so in that case water is used as multiplier of force to lift a really heavy piece of object.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yet the pressure acting in the water is same across all objects. Is it correct to say that water is dispersing the required force all through the water hence reducing the force required to lift the vehicle?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is such a fascinating concept.. Quite the counterintuitive nature of our nature.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that it is correct to say that the effect of a force, is the production of the correspondig pressure inside the fluid which is contacting that force, and in turn the effect of a pressure of a fluid, is the force produced on all the surfaces which are contacting that fluid

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1and causeeffect, namely forcepressure, are descrined by the subsequent formula: \[\Large P = \frac{F}{S}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And depending on the area of surface that force induced by the object contacting the surface leading to space filled with liquids without leaks connected to another surface counteracting the force will be all dependent on the area upon which the respective forces are taking effect.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So within the given pressure, greater surface implies more force required to meet the same pressure, and less surface means less force required to meet that pressure.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1no, greater is the surface and greater is the forces, and less is the surface and less is the force, since the ratio force/surface has to be constant. The magnitude of the resultant forces, or of the applied forces depends on the area of the surfaces contacting those forces

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah I realized I said the wrong

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! your last reply is correct!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isn't this counterintuitive?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, I guess the expanse of liquid present underneath the surface with greater area will make sense that corresponding force required induce such pressure is indeed greater

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On the other hand smaller surface with smaller liquid underneath would make it less force required the meet the pressure although distributed throughout all the fluid bodies.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I don't think, I think that it is intuitive, the problem is that we have no experience, in our daily life, about expansion of fluid

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes! Yes! lack of experience with interaction of liquid makes it harder for us to visualize.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! and the best way is to use our imagination. As Einstein said: "Imagination is more important than knowledge"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0On the other hand, studying physics makes you capable of physically achieving anything if you know how to properly coordinate your muscles to execute the proper force required to achieve that phenomena.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think, in my personal opinion, all children should be taught the principles of physics so as to excel as well as increase "efficiency" in life. So much are wasted by ignorance

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think too, that physics offers us a exemplary thinking mechanism.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Also physics extends our lifespan

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By creating new neurons in your brain

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1I think that there is a time for study physics, what is important for children is play, since by means of play, all children can collect those experience which will allow to think in terms of our imagination

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0True, from those natural experience will help us to relate physics to our past times. Richer the past times, greater the utility for physics

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And physics makes us nostalgic of our past memories.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And ultimately makes life worthwhile. Physics makes all my struggles disappear.

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1throw a rock in the air, or running through a meadow, are all important things

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Physics is the best prescription for eliminating societal prejudice and personal stigmas for it provides us more objective insight upon which to rely all of the issues we encounter in life. This makes it easy for us to say that someone is wrong in saying something or the applications of medical physics such as how our neutrons interact would invent better ways to learn

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it's all physics after all our world is made of. And contemporary education has so far failed to quantify the most effective means by which to learn, hence resulting in "learning something until it finally sinks in" or just relying on our intuition rather than physics of neurology.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's why I love physics .

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I bet you were on the same road when you were younger

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1not exactly, when I was younger, I had the concern, to give a logical accommodation to everything around me

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I see, trying to explain mathematically everything around you o

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! a logical or rational explanation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1ok! when you need help in physics, keep in mind that I always will help you! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks so much Michele!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you mind proof reading another application from this question that I learned?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02. I calculated the pressure acting on the floor when standing on both feet. The areas of my shoes were approximated to be 15cm^2 each, making ti 30cm^2. Given that I weigh around 58kg, and pressure is calculated as P=Mg/s (surface area) P=58kg*9.8m/s^2/0.003m^3=189466.667 The pressure acting on the floor when standing on both feet is 1.9*10^5 Pa

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0According to your P=Mg/s I think my reasoning above seems pretty good

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1yes! Your reasoning is correct!
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